Stress and strain of cracked sections - Concrete Structures: Stresses and Deformations - page 236

Example 7.2 Cracked T section subjected to M and N
Solve Example 7.1, assuming that the section is subjected to a bending
moment of 1000 kN-m (8850 kip-in) and a normal force of − 800 kN

( 180 kip) at a point 1.0 m (40 in) below the top edge of the section. The

cross-section dimensions and moduli of elasticity of steel and concrete
are the same as in Example 7.1 (Fig. 7.7(a) ).
The resultant force on the section is a normal force of

−

− 800 kN at a

distance 0.25 m above the top edge. Thus, e s 1.45 m.

Substituting in Equation (7.20) and solving for c , the height of the
compression zone, gives:

= − (0.25 + 1.20) = −

c = 0.444 m (17.5 in).

The e ective area is shown in Fig. 7.7(c). The transformed section is

composed of the area of concrete in compression plus

ff

α ( A s + A ′ s ) with α

= 6.667. The distance between point O, the centroid of the

transformed section, and the top edge is calculated to be y

200/30 =

0.229 m
(Fig. 7.7(c) ). The area and moment of inertia of the transformed
section about an axis through its centroid

=

A = 0.3073 m 2 I = 31.73 × 10 −3 m 4 .

If A ′ s is ignored, Tables 7.1, 7.3 and 7.4 may be used, giving:

c = 0.46 m y = 0.24 m I = 30 × 10 −3 m 4 .

Transform the given bending moment and normal force into an
equivalent system of a normal force N at the centroid of the trans-
formed section combined with a bending moment M .

N = − 800 kN

M = 1000 × 10 3 − 800 × 10 3 (1.000 − 0.229)

= 383.2 kN m (3400 kip in).

The strain at O and the curvature (Equation (2.16) )

800×10 3
0.3073 30×10 9 −

1

ε O = = − 87 × 10 −6